# Maths Quiz on Differential Equations for IIT JEE & Engineering Exam

0 ## JEE Maths Quiz on Differential Equations

JEE Maths Quiz on Differential Equations : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Differential Equations.

## Quiz on Differential Equations

Q1.  If x dy = y (dx + y dy), y > 0 and y (1) = 1, then y (−3) is equal to _________.
a) 0
b) 3
c) 5
d) -3

b) 3

Solution:

x dy = y (dx + y dy)

[x dy − ydx] / [y2] = dy

−d (x / y) = dy

Integrating both sides, we get x / y + y = c [Because y (1) = 1 ⇒ c = 2; Hence xy + y = 2 ]

For x = −3,

y− 2y − 3 = 0

⇒ y = −1 or 3

⇒y = 3 (Because y > 0)

Q2.  (x+ y2) dy = xy dx. If y (x0) = e, y (1) = 1, then the value of x0 = __________.
a) ± √2e
b) ± √3e
c) ± √4e
d) ± √5e

b) ± √3e

Solution:

(x+ y2) dy = xy dx

x (x dy − y dx) = −y2 dy

[x * (y dx − x dy)] / y2 = dy

[x / y] d (x / y) = dy / y

Integrating, x/ 2y2 = loge y + c

Given y (1) = 1

c = 1 / 2

x/ 2y= loge y + 1 / 2

Now y (x0) = e

x0/ 2e− loge e − 1 / 2 = 0

x0= 3e2

x= ± √3e

Q3. Solution of differential equation 2xy dy / dx = x2 + 3y2 is __________.
a) x2 + y= px2
b) x2 – y= px3
c) x3 + y3 = px2
d) x2 + y= px3

d) x2 + y= px3

It is homogeneous equation dy / dx = [x2 + 3y2] / 2xy

Put y = vx and dy / dx = v + x * [dv / dx]

So, we get x * [dv / dx] = [1 + v2] / 2v

2v dv / 1 + v2 = dx / x

On integrating, we get

x2 + y= px3. (where p is a constant)

Q4.  An integrating factor for the differential equation (1 + y2) dx − (tan−1 y − x) dy = 0
a) etan−1y
b) etan+1y
c) etan−2y
d) etan+2y

a) etan−1y

Solution:

(1 + y2) dx − (tan−1 y − x) dy = 0

dy / dx = (1 + y2) / (tan−1 y − x)

dx / dy = [tan−1 y / 1 + y2] − [x / 1 + y2] [dx / dy] + [x / 1 + y2] = [tan−1 y] / [1 + y2]

This is equation of the form dx / dy + Px = Q

So, I.F. = e∫Pdy

= e∫1 / 1 + y2.dy

= etan−1y

Q5. What is the general solution of the differential equation (2x − y + 1) dx + (2y − x + 1) dy = 0?

Solution:

(2x −y + 1) dx + (2y − x + 1) dy = 0

dy / dx = 2x − y + 1x − 2y −1, put x = X + h, y = Y + k

dY / dX = [2X − Y + 2h − k + 1] / [X − 2Y + h −2k − 1]

2h − k + 1 = 0

h − 2k − 1 = 0

On solving h = −1, k = −1;

dY/ dX = [2X − Y] / [X − 2Y]

Put Y = vX;

dY / dX = v + [X dv / dX]

v + [X dv / dX] = [2X − vX] / [X − 2vX] = [2 − v] / [1 − 2v]

X dv / dX = [2 − 2v + 2v2] / [1 − 2v] = 2 (v2 −v + 1) / [1 − 2v]

dX / X = (1 − 2v) / 2 (v− v + 1) dv

Put v− v + 1 = t

(2v − 1) dv = dt

dX / X = −dt / 2t

log X = logt− ½ + log c

X = t−1/2 c

X = (v2 − v + 1)−1/2 * c

X(v2 − v + 1) = constant

(x + 1)(([y + 1)/ (x + 1)2] − [(y+1) / (x + 1)] + 1) = constant

(y + 1)− (y + 1) (x + 1) + (x + 1)= c

y2 + x2 − xy + x + y = c

Q6.  If y′= [x − y] / [x + y], then its solution is ____________.

Solution:

Given dy / dx = x −y / x + y.

Put y = vx

dy / dx = v + x * [dv / dx]

v + x * [dv / dx] = [x − vx] / [x + vx]

v + x [dv / dx] = [1 − v] / [1 + v] [1 + v] / [2 − (1 + v)2] dv = dx / x

Integrating both sides, ∫[1 + v] / [2 − (1 + v)2] dv = ∫dx / x

Put (1 + v)= t

⇒ 2 (1 + v) dv = dt

[1 / 2] ∫dt / [2 − t] =∫dx / x

[− 1 / 2] log (2 − t) = log xc

[−1 / 2] log [2 − (1 + v)2] = log xc

[−1 / 2] log [−v− 2v + 1] = log xc

log 1 / √[1 − 2v −v2] = log xc

x2c(1 − 2v −v2) = 1

y2 + 2xy − x2 = c1.

Q7. A function y = f (x) has the second-order derivatives f′′(x) = 6 (x − 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x − 5, then the function is __________.
a) (x + 1)3
b) (x − 1)2
c) (x − 1)3
d) (x + 1)2

c) (x − 1)3

Solution:

Given f′′(x) = 6 (x − 1)

f′(x) = 3 (x − 1)+ c1 ..(i)

But at point (2, 1) the line y = 3x − 5 is tangent to the graph y = f(x).

Hence, dy / dx∣x = 2 = 3 or

f′ (2) = 3.

Then from (i) f′ (2) = 3 (2 − 1)+ c1

3 = 3 + c1

c= 0 i.e.,

f′ (x) = 3 (x − 1)2

Given f (2) = 1

f (x) = (x − 1)+ c2

f (2) = 1 + c2

1 = 1 + c2

c= 0

Hence, f (x) = (x − 1)3.

Q8. The general solution of the differential equation (x + y) dx + x dy = 0 is _______.

Solution:

(x + y) dx + x dy = 0

x dy = − (x + y) dx

dy / dx = −[x + y] / x

It is homogeneous equation, hence, put y = vx and dy / dx = v + x [dv / dx], we get

v + x dv / dx = −[x + vx] / x= −[1 + v] / 1

x dv / dx = −1 −2v

∫dv / [1 + 2v] = −∫dx / x

[1 / 2] log (1 + 2v) = −log x + log c

log (1 + 2 [y / x]) = 2 log [c / x] [x + 2y] / x = (c / x)2

x+ 2xy = c.

Q9. Equation of curve through the point (1, 0) which satisfies the differential equation (1 + y2) dx − xy dy = 0, is ________.
a) 3
b) 5
c) 2
d) 1

d) 1

We have [dx / x] = [y dy] / [1 + y2]

Integrating, we get log |x| = ([1 / 2] * log [1 + y2]) + log c

|x| = c √(1 + y2)−−−−−−−

But it passes through (1, 0), so we get c = 1

Therefore, the solution is x2 = y2 + 1 or x2 − y= 1.

Q10. The equation of the curve which passes through the point (1, 1) and whose slope is given by 2y / x, is __________.
a) x2
b) 2x2
c) 1/2x2
d) x3

a) x2

Slope dy / dx = 2y / x

2 ∫dx / x =∫dy / y

2 log x = log y + log c

x= yc

Since it passes through (1, 1), therefore, c = 1.

Hence, x− y = 0

y = x2.

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